objective-c 将字符转换为键盘码的代码
2018-07-20 来源:open-open
- (int)keyCodeForCharacter: (NSString*)character { if(![character length]) return -1; char code; BOOL shift, alt; if(Ascii2Virtual( (char)[character characterAtIndex: 0],shift,alt,code)) { return code; } return -1; } BOOL Ascii2Virtual(char pcar, BOOL *pshift, BOOL *palt, char *pkeycode) { KeyboardLayoutRef keyboard; const void *keyboardData; // keyboard layout data UInt16 nbblocs; char *modblocs, *blocs, *deadkeys; int ix, ifin, numbloc, keycode; BOOL shift, alt; // get the current keyboard if(KLGetCurrentKeyboardLayout(&keyboard;)) return NO; // get the description of the current keyboard layout if(KLGetKeyboardLayoutProperty(keyboard, kKLKCHRData, &keyboardData;)) return NO; // get pointer early numbers of blocks for each combination of modifiers modblocs = ((char *)keyboardData) + 2; // get number of blocks keycode->ascii nbblocs = *((UInt16 *)(keyboardData + 258)); // get pointer early blocks keycode-> ascii blocs = ((char *)keyboardData) + 260; // determining the size of all tables keycode-> ascii a scanner ifin = nbblocs*128; // determining pointer early in the tables of dead keys deadkeys = blocs+ifin; // Now it runs blocks keycode-> ascii to find the car ascii for (ix=0; ix<ifin ; ix++) { if (blocs[ix]==pcar) { // found ascii value: now we must determine which block it is keycode = ix & 0×7f; // 0111 1111 mask numbloc = ix >> 7; break; } } // not found: bail out (error) if (ix >= ifin) return NO; // from block number, we must find the combination of modifiers using this block for (ix=0; ix<15; ix++) { // it does not address whether the modifiers are not "capital" and "option" if (ix&1 || ix&4) continue; // Combining modifiers found for the block if (modblocs[ix]==numbloc) { shift = (ix&2) ? YES : NO; alt = (ix&8) ? YES : NO; break; } } // combination modifiers not found: bail if (ix>=15) return NO; // save our parameters *pkeycode=keycode; *pshift=shift; *palt=alt; return YES; }
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